Help with $\lim_{x\rightarrow +\infty} (x^2 - \sqrt{x^4 - x^2 + 1})$

Question

$\lim_{x\rightarrow +\infty} (x^2 - \sqrt{x^4 - x^2 + 1}) = ?$

I don't know how to solve the indetermination there... is it possible to rearrange the expression in brackets in order to use L'Hospital or Taylor Series?

Answer 1

Hint: $$x^2-\sqrt{x^4-x^2+1}=\left(x^2-\sqrt{x^4-x^2+1}\right)\cdot\frac{x^2+\sqrt{x^4-x^2+1}}{x^2+\sqrt{x^4-x^2+1}}=\frac{(x^2)^2-\left(\sqrt{x^4-x^2+1}\right)^2}{x^2+\sqrt{x^4-x^2+1}}$$

Once you have that simplified, multiply by $\dfrac{1/x^2}{1/x^2}$ and recognize that $x^2=\sqrt{x^4}$ to distribute the $1/x^2$ 'into' the radical.

Answer 2

You may write $$ x^2 - \sqrt{x^4 - x^2 + 1}=\frac{(x^2)^2 - (x^4 - x^2 + 1)}{x^2 + \sqrt{x^4 - x^2 + 1}}=\frac{1 - \frac{1}{x^2}}{1 + \sqrt{1 - \frac{1}{x^2} + \frac{1}{x^4}}} \to \frac12 $$ as $x \to +\infty$.

Answer 3

I prefer to set the limit $\to0$ which often eases of calculation

Set $\dfrac1{x^2}=h\implies h\to0^+\implies h>0$

$\lim_{x\rightarrow +\infty} (x^2 - \sqrt{x^4 - x^2 + 1}) $

$$=\lim_{h\to0^+}\frac{1-\sqrt{1-h+h^2}}h$$

$$=\lim_{h\to0^+}\frac{1-(1-h+h^2)}{h(1+\sqrt{1-h+h^2})}=\cdots$$